What is the equation of the parabola that has a vertex at # (38, -22) # and passes through point # (3,7) #?

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Answer 1 · 2017-07-17T01:03:07

There are 2 standard equations:

#y = 29/1225(x-38)^2-22#

#x = -35/841(y-(-22))^2+38#

Actually, there are two such parabolas, one whose general vertex form is:

#y = a(x-h)^2+k" [1]"#

and the other whose general vertex form is:

#x = a(y-k)^2+h" [2]"#

where #(h,k)# is the vertex and "a" is found by substituting in the given point #(x,y)#

Substitute the vertex #(38,-22)# into equations [1] and [2]:

#y = a(x-38)^2-22" [1.1]"#

#x = a(y-(-22))^2+38" [2.1]"#

Substitute the point #(3,7)# into equations [1.2] and [2.1] and then solve for "a":

#7 = a(3-38)^2-22# and #3 = a(7-(-22))^2+38#

#29 = a(3-38)^2# and #-35 = a(7-(-22))^2#

#29 = a(-35)^2# and #-35 = a(29)^2#

#a = 29/1225# and #a = -35/841#

Substitute the values for "a" into its respective equation:

#y = 29/1225(x-38)^2-22" [1.2]"#

#x = -35/841(y-(-22))^2+38" [2.2]"#

Here is a graph of the two points and both parabolas:

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